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\author{Big Stupid Melon \\ 41304296 }
\title{Solutions to the Homework}

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\section*{Problem 1}

\subsection*{(a) Derive the unique risk-neutral probability measure.}

The risk-neutral probability measure is 

$$ p_u = \frac{e^{rT} - d}{u - d} =  0.6024$$
$$ p_d = 1 - q_u = 0.3976 $$


\subsection*{(b) Price the call option using the main asset pricing formula.}

The pay off of the option is $f_u = q_{1u} - K = 40$, $f_d = 0$.
Thus the price of the European call can be calculated as

$$ q_c = e^{-rT} (f_u p_u + f_d p_d) = \$24.05 $$


\subsection*{(c) Hedging for the put option.}

First, find the hedging portfolio of the European put. Assume that this portfolio consists of $a_c$ shares of the stock and $\beta_c$ shares of the bond. Then we have:

$$  0 =  a_c \cdot 80 + \beta_c \cdot 1.002 $$
$$ 10 =  a_c \cdot 30 + \beta_c \cdot 1.002 $$

We can get $a_c = -0.2$, $\beta_c = 15.9681$. By no-arbitrage pricing theorem, we can calculate the price of the European put as

$$ q_p = 60 a_c + \beta_c = \$3.97 $$


\section*{Problem 2}

\subsection*{(a) EPV of immediate investment}

Write the EPV of immediate investment as a function of C.

\begin{eqnarray*}
V_{0, 0} &=& 0.6 \cdot \sum_{t = 1}^{\infty} \frac{110 - 1.1C}{1.1^t} + 0.4 \cdot \sum_{t = 1}^{\infty} \frac{110 - 0.9C}{1.1^t} - 335 \\
&=& 0.6 \cdot  \frac{110 - 1.1C}{0.1} + 0.4 \cdot  \frac{110 - 0.9C}{0.1} - 335 \\
&=& 765 - 10.2C
\end{eqnarray*}
provided $C < 75$. The entrepreneur will invest $0$ if $C \ge 75$. Thus
$$ V_{0, 0}(C)=\left\{
\begin{aligned}
&765 - 10.2C  &,  \  C < 75 \\
&0  &,   \ C \ge 75
\end{aligned}
\right.
$$


\subsection*{(b) EPV of investment at $t = 1$}

Similarly, we can obtain
$$ V_{1, 1}(1.1C) =   \sum_{t = 1}^{\infty} \frac{110 - 1.1C}{1.1^t}  - 335 = \left\{
\begin{aligned}
&765 - 11C   &,  \  C < 69.55 \\
&0  &,   \ C \ge 69.55
\end{aligned}
\right.$$
$$ V_{1, 1}(0.9C) =   \sum_{t = 1}^{\infty} \frac{110 - 0.9C}{1.1^t}  - 335 = \left\{
\begin{aligned}
&765 - 9C   &,  \  C < 85 \\
&0  &,   \ C \ge 85
\end{aligned}
\right.$$

Hence,
$$ V_{1, 1}(C) = \left\{
\begin{aligned}
&\frac{765 - 10.2C}{1.1}   &, & \  C \le 69.55 \\
&\frac{0.4(765 - 9C )}{1.1} &,& \  69.55 < C < 85 \\
&0  &, &  \ C \ge 85
\end{aligned}
\right.$$


\subsection*{(c) Optimal investment strategy}

Find a $C^*$ solves the following equation:

$$765 - 10.2C \ge \frac{0.4(765 - 9C )}{1.1}$$
We get $C^* \le 70.28 $. Then it's easy to verify that:
\begin{itemize}
\item if $C \le 70.28$, invest immediately.
\item if $C \in (70.28, 85)$, do not invest at $t = 0$; invest at $t = 1$ only if the operating cost drops, otherwise never invest. 
\item if $C \ge 85$, don't invest.
\end{itemize}


\section*{Problem 3}

The present value of the project is
$$ V_0 = \sum_{\omega \in \Omega} p(\omega)V(\Omega) = 0.5\cdot \frac{100}{0.1}  + 0.6 \cdot \frac{0}{0.1} = 500$$

If the manager decided to invest at $t = 1$, for any $I < 0$ she will not invest if $S_1 = 40$(because it will lead to zero profit). If $S_1 = 60$, then the present value of the project at time $t = 1$ is 1000. obviously, if $I > 1000$, the manager will never invest. The EPV of the project if the manager invest at $t = 1$ is

$$ V_{0,1} = \beta(0.5(1000 - I)) = \frac{500 - 0.5I}{1.1}$$

At time $t = 0$, we have $V_0 = 500$, which means for any $I > 500$, the manager will not choose to invest at $t = 0$. Find the value $I^*$ which makes the manager's investing at $t = 0$ dominates investing at $t = 1$:
$$500 - I^* \ge \frac{500 - 0.5I^*}{1.1} $$
We can get $I^* \le \frac{250}{3}$.

To conclude, the manager invest at
\begin{itemize}
\item $t=0$, if $I \le \frac{250}{3}$.
\item $t = 1$, if $S_1 = 60$ and $\frac{250}{3} < I < 1000$.
\item never invest, if $I \ge 1000$.
\end{itemize}


\section*{Problem 4}

At time $t = 2$, if the price rises, the value of the firm will be
$$V_2(50) = \sum_{t = 1}^{\infty} \beta^t (R_2 - C) = \frac{250 - 200}{0.1} = 500$$
Here we assume that scrap value $Sc^h$ is no more than \$500 mln.

If the price drop to 30, then the firm will be closed. Clearly,
$$V_2(30) = -50 + \frac{Sc^l}{1.1} $$

The time-1 EPV reads
$$ V_1(40) = E^{\mathbb{Q}} [\beta V_2] = \frac{1}{1.1}(0.5 \cdot 500 + 0.5  \cdot (-50 + \frac{Sc^l}{1.1}))$$
Find a $Sc^{l*}$ to make it better to disinvest at time $t = 1$
$$ V_1(40) < NPV_1(40) = \frac{150}{1.1} $$
And we get $Sc^{l*} = -165 $. Considering the limited liability constrain, the manager will not choose to disinvest at time $t = 1$. If $Sc^h \ge 500$, for any $0 \le Sc^l < Sc^h $, she will disinvest at $t = 2$.


\section*{Problem 5}

In this case, the EPV of the future profit can be write as
$$E_1^\mathbb{Q}[\beta V_2^{ac}] = E^\mathbb{Q}[\beta V_2^{ac} | S_1 = 40] = 0.4 \cdot 500 + 0.6 \cdot \frac{313.64}{1.1} = 371.08$$

$$V_0^{ac} = E^\mathbb{Q}[\beta V_1^{ac}] = 0.5 \cdot 1000 + 0.5 \cdot \frac{371.08}{1.1} = 668.67$$

At time $t = 2$, there's no uncertainty, everything is the same to the lecture note.

At time $t = 1$, if $S_1 = 60$, similarly, it's better to invest at time $t = 1$. If $S_1 = 40$, if the manager invest, she will gain $371.08 - 500 < 0$, thus she won't invest.

At time $t = 0$, the manager calculate the EPV of the investment at time t = 1 or later as
$$\beta E^\mathbb{Q}[G_1] = \frac{1}{1.1} [0.5 \cdot 500 + 0.5\cdot 0] = 227.27$$
If she invest now, she will gain
$$V_0^{ac} - I = 668.67 - 500 = 168.67 < 227.27$$

To summarize,
\begin{itemize}
\item Do not invest at time $t = 0$.
\item Invest at time $t = 1$ if $S_1 = 60$.
\item Invest at time $t = 2$ if $S_2 = 50$.
\end{itemize}

It's the same to the lecture note, the intuition is when we move some probability from $S_2 = 50$ to $S_2 = 30$, it actually made the invest environment worse.


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